We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

1. The radioisotope Cobalt-60 is used in cancer therapy. The half-life of this isotope is 5.27 years.

Which equation determines the percent of an initial amount of the isotope remaining after t years?

A=5.27(12)100tA=5.27(12)100t

A=100(12)t5.27A=100(12)t5.27

A=5.27(12)t100A=5.27(12)t100

A=100(12)5.27t

**My Answer-'C'**

2. What are the real and imaginary parts of the complex number?

9+7i

The real part:

The imaginary part:

**My Answer- Real part=9, Imag. part=7i**

3.

What is the simplest radical form of the expression?

(x4y7)23

xy2xy−−√3xy2xy3

x6y10y√x6y10y

x2y4x2y2−−−−√3x2y4x2y23

x3y5x√

**My Answer-'D'**

**Thanks so much**

Guest Jan 16, 2018

#1**+1 **

The correct equation in the first part should be

A = 100(1/2)^{t/5.27}

To see that this is correct....let t = 5.27 and we have that

A = 100(1/2)^{ 5.27/5.27} = 100 (1/2)^{1} = 100 (1/2) = 50% left after 5.27 years, the half-life

Your second answer is correct

I can't tell exactly what the third one is....sorry....!!!

CPhill Jan 16, 2018